Relations --- Binary Relations. Debdeep Mukhopadhyay IIT Madras

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1 Relations --- Binary Relations Debdeep Mukhopadhyay IIT Madras

2 What is a relation? The mathematical concept of relation is based on the common notion of relationships among objects: One box is heavier than the other One man is richer than the other An event occurs prior to the other

3 Ordered n-tuple For n>0, an ordered n-tuple (or simply n- tuple) with ith component a i is a sequence of n objects denoted by <a 1,a 2,,a n >. Two ordered n-tuples are equal iff their ith components are equal for all i, 1<=i<=n. For n=2, ordered pair For n=3, ordered triple

4 Cartesian Product Let {A 1,A 2,,A n } be an indexed collection of sets with indices from 1 to n, where n>0. The cartesian product, or cross product of the sets A 1 through A n, denoted by A χ 1 A χ 2 χ A n, or n i=1 Ai is the set of n-tuples <a 1,a 2,,a n > a i ЄA i }. n When A i =A, for all i, then i=1 Ai will be denoted by A n.

5 Examples Let A={1,2}, B={m,n}, C={0}, D=Φ. AxB={<1,m>,<1,n>,<2,m>,<2,n>} AxC={<1,0>,<2,0>} AxD= Φ When A and B are real numbers, then AxB can be represented as a set of points in the Cartesian Plane. Let, A={x 1 x 2} and B={y 0 y 1}. Then AxB={<x,y> 1 x 2 Λ 0 y 1}

6 Theorems 1. AX(B U C)=(AXB) U (AXC) 2. AX(B C)=(AXB) (AXC) 3. (A U B)XC=(AXC) U (BXC) 4. (A B)XC=(AXC) (BXC)

7 Proof of 1 < xy, > A ( B C) x A y ( B C) x A ( y B y C) ( x A y B) ( x A y C) ( < xy, > A B) ( < xy, > A C) < xy, > ( A B) ( A C) The rest of the proofs are similar.

8 What is a relation mathematically? Let A 1,A 2,,A n be sets. An n-ary relation R on n n i=1 Ai is a subset of i=1 Ai. If R=Φ, then R is called the empty or void relation. If A n R= i=1 i then R is the universal relation. If A i =A for all i, then R is called an n-ary relation on A. If n=1, unary If n=2, binary Ternary

9 Number of n-ary relations n If A i has r i elements, then i=1 Ai has rielements i= 1 The number of n-ary relations is the cardinal number of the power set of the cartesian product of the A i s. Thus, the number of relations is n n i= 1 2 r i

10 Equality of relations n Let R 1 be an n-ary relation on i=1 Ai and R 2 be an m m-ary relation on i=1bi. Then R 1 =R 2 iff n=m, and A i =B i for all i, 1 i n, and R 1 =R 2 are equal sets of ordered n-tuples. Every n-ary relation on a set A, corresponds to an n-ary predicate with A as the universe of discourse. A unary relation on a set A is simply a subset of set A.

11 Binary Relations

12 They are frequently used in abstraction in CS Various data structures, like trees and graphs can be modeled as binary relations and vice versa. We shall see techniques and methods to analyze.

13 Binary Relations Let A, B be any two sets. A binary relation R from A to B, written (with signature) R:A B, is a subset of A B. E.g., let < : N N : {<n,m> n < m} The notation a R b or arb means <a,b> R. E.g., a < b means (a,b) < If arb we may say a is related to b (by relation R), or a relates to b (under relation R). A binary relation R corresponds to a predicate function P R :A B {T,F} defined over the 2 sets A,B; e.g., eats : {<a,b> organism a eats food b}

14 Domain and Co-domain Let R be a binary relation over AxB. Domain: Set A Co-domain: Set B <a,b>єr=> arb <a,b> Є R => arb

15 Complementary Relations Let R:A B be any binary relation. Then, R:A B, the complement of R, is the binary relation defined by R : {<a,b> (a,b) R} Example: < = {(a,b) (a,b) <}} = {(a,b) ) a<b} } =

16 Inverse Relations Any binary relation R:A B has an inverse relation R 1 :B A, defined by R 1 : {(b,a) (a,b) R}. E.g., < 1 = {(b,a) a<b} = {(b,a) b>a} = >. E.g., if R:People Foods is defined by arb a eats b, then: br 1 a b is eaten by a. (Passive voice.)

17 Relations on a Set A (binary) relation from a set A to itself is called a relation on the set A. E.g., the < relation from earlier was defined as a relation on the set N of natural numbers. The identity relation I A on a set A is the set {(a,a) a A}.

18 Representing Relations With a zero-one matrix. With a directed graph.

19 Using Zero-One Matrices To represent a relation R by a matrix M R = [m ij ], let m ij = 1 if (a i,b j ) R, else 0. E.g.,A={1,2,3}, B={1,2}. Let R be the relation from A to B containing (a,b) s.t a is in A and b is in B and a>b. The 0-1 matrix representation 0 0 M R = When A=B, we have a square matrix

20 So, what is complement of R? A={1,2,3}, B={1,2}. Let R be the relation from A to B containing (a,b) s.t a is in A and b is in B and a>b Complement of R = {(a,b) not(a>b)} ={(a,b) a<=b} 0-1 matrix is: M R 1 1 = We can obtain by the element wise bit complement of the matrix.

21 Types of Relations Let R be a binary relation on A: R is reflexive if xrx for every x in A R is irreflexive if xrx for every x in A R is symmetric if xry implies yrx for every x,y in A R is antisymmetric if xry and yrx together imply x=y for every x,y in A R is transitive if xry and yrz imply xrz for every x,y,z in A

22 Zero-One Reflexive, Symmetric These relation characteristics are very easy to recognize by inspection of the zero-one matrix anything anything anything anything anything anything Reflexive: all 1 s on diagonal Irreflexive: all 0 s on diagonal Symmetric: all identical across diagonal Antisymmetric: all 1 s are across from 0 s

23 Tell what type of relation M R M M R R = (Reflexive, Symmetric) = (Neither Reflexive nor irreflexive, Symmetric) = (Reflexive, Anti-Symmetric) 0 0 1

24 Operations on 0-1 Matrix Union and Intersection of relations can be obtained by join and meet of the Binary matrices M = M M R R R R M = M M R R R R

25 Operations on 0-1 Matrix M M R = M R = = M = R1 R2 R1 R2

26 Composition of relations R: A B, S:B C S R : A C Suppose, A, B and C have m, n and p elements M S :[s ij ] (nxp), M R :[r ij ](mxn), M S.R :[t ij ](mxp) (a i,c j ) belongs to S.R iff there is (a i,b k ) belonging to R and (b k,c i ) belonging to S for some k. Thus t ij =1 iff r ik =1 and s kj =1, for some k. Thus, M = M M SR R S

27 Example of composition MR = M S = MSR MR M S = = 0 0 0

28 Using Directed Graphs A directed graph or digraph G=(V G,E G ) is a set V G of vertices (nodes) with a set E G V G V G of edges (arcs,links). Visually represented using dots for nodes, and arrows for edges. Notice that a relation R:A B can be represented as a graph G R =(V G =A B, E G =R). M R 0 0 = Edge set E G (blue arrows) 1 2 Node set V G (black dots)

29 Digraph Reflexive, Symmetric It is extremely easy to recognize the reflexive/irreflexive/ symmetric/antisymmetric properties by graph inspection. Reflexive: Every node has a self-loop Irreflexive: No node links to itself Symmetric: Every link is bidirectional Antisymmetric: No link is bidirectional

30 A Question discussed in class Does symmetricity and transitivity imply reflexivity? Reason of doubt: arb=>bra (symmetricity) This implies ara (transitivity) So, R is reflexive! Clarification: Symmetric, Transitive But not Reflexive..

31 Closure of Relations

32 Closure? Let R be a relation on a set A R may or may not have a property P Define S, as the relation which has the property P AND S contains R AND S is the subset of every relation with property P and which contains R S is called the closure of R w.r.t P Closure may not exist.

33 Reflexive Closure R={(1,1),(1,2),(2,1),(3,2)} on the set A={1,2,3} Is R reflexive? How can we create an S (which is as small as possible) containing R which is reflexive? Add (2,2) and (3,3). S is reflexive and contains R Since, any reflexive relation on A must contain (2,2) and (3,3), all such relations must be a superset of S S is hence the reflexive closure.

34 Generalization Define ={(a,a) a Є A} (Diagonal Relation) S=R U S is the reflexive closure of R.

35 Symmetric Closure R={(1,1),(1,2),(2,2),(2,3),(3,1),(3,2)} on the set A={1,2,3} Is R symmetric? How can we create an S (which is as small as possible) containing R which is symmetric? Add (2,1) and (1,3). S is symmetric and contains R Since, any symmetric relation on A must contain (2,1) and (1,3), all such relations must be a superset of S S is hence the symmetric closure.

36 Generalization Define R -1 ={(b,a) (a,b)єr} R={(1,1),(1,2),(2,2),(2,3),(3,1),(3,2)} R -1 ={(1,1),(2,1),(2,2),(3,2),(1,3),(2,3)} S=R U R -1 ={(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2)} S contains R All such relations contain S Thus, S is the symmetric closure.

37 Transitive Closure? R={(1,3),(1,4),(2,1),(3,2)} R is not transitive. So, add (1,2),(2,3),(2,4),(3,1). Does it become transitive? No, because say (3,2) and (2,4) are members but not (3,4). So, transitive closure is not that easy.

38 Composition of R with itself : R n Let R be a relation on set A arb => {(a,b) (a,b)єr} Let R be a relation on the set A. The powers R n, n=1, 2, 3, are defined recursively by: R 1 =R and R n+1 =R n.r Example: R={(1,1),(2,1),(3,2),(4,3)} R 2 ={(1,1),(2,1),(3,1),(4,2)}

39 Composition in DAG A path from a to b in DAG G, is a sequence of edges (a,x 1 ),(x 1,x 2 ),,(x n-1,b). The path has length n. A path of length n 1 that begins and ends at the same vertex is called a circuit or cycle. Theorem: Let R be a relation on a set A. There is a path of length n, where n is a positive integer from a to b, iff (a,b) belongs to R n.

40 Proof Base: There is a path from a to b of length 1, iff (a,b) is in R. Induction: Assume theorem is true for n There is a path of length (n+1) between a and b, iff there is a path of length 1 between (a,c) and there is a path of length of n between (c,b) for some c. Hence, there is such a path iff (a,c)єr and (c,b) ЄR n (inductive hypothesis) But there is such an element c iff (a,b) ЄR n+1

41 Theorem The relation R on a set A is transitive iff n R R If part: If R 2 is a subset of R (special case) then R is transitive Else part: Trivial proof for n=1 Assume if R is transitive R n is a subset of R. Consider (a,b)єr n+1. Thus, there is an element c st (a,c) Є R and (c,b) Є R n. By hypothesis, (c,b) Є R. But R is transitive, so (a,c) Є R and (c,b) Є R means (a,b) Є R

42 Now lets look at the Problem of Transitive Closure Define, the connectivity relation consisting of the pairs (a,b) such that there is a path of length at least one from a to b in R. R + = R n= 1 Theorem: The connectivity relation is the transitive closure Proof: R + contains R R + is transitive n

43 To show that R + is the smallest! Assume a transitive S containing R R + is a subset of S + (as all paths in R are also paths in S) Thus, we have + + n R S S (as S is transitive we have S S)

44 Lemma Let, A be a set with n elements, and let R be a relation on A. If there is a path of length at least one in R from a to b, then there is such a path with length not exceeding n. Thus, the transitive closure is tr ( ) n = R i= 1 i Proof follows from the fact R k is a subset of t(r)

45 Example Find the zero-one matrix of the transitive closure of M R = M = M M M R R [2] [3] R R R = = 2 3 M 0 1 0, M M R + + R = =

46 Algorithm-1 Procedure transitiveclosure(m R ) A=M R, B=A for i=2 to n begin A = A B B = B A end B is the answer Complexity: n 2 (2n-1)(n-1)+n 2 (n-1)=o(n 4 )

47 Algorithm-2 (Roy-Warshall algorithm) Based on the construction of 0-1 matrices, W 0,W 1,,W n, where W 0 =M R (0-1 matrix of the relation). Uses the concept of internal vertices of a path: If there is a path (a,b), namely, (a,x 1,x 2, x m-1,b) Internal vertices: x 1,x 2, x m-1 The start vertex is not an internal vertex unless it is visited again, except as a last vertex The end vertex is not an internal vertex unless it has been visited before, except as a first vertex

48 So, what is the trick? Construct, W k =[w (k) ij ], where w (k) ij =1, if there is a path from v i to v j such that all the interior vertices of this path are in the set {v 1,v 2,,v k }, and 0 otherwise. W n =M R*. Can you see why? But construction of W n is easy than the boolean product of matrices.

49 Construct W k w ij (k) =1, if there is a path from v i to v j such that all the interior vertices of this path are in the set {v 1,v 2,,v k }, and 0 otherwise. CASE-1 v i Internal Nodes from set {v 1,v 2,,v k-1 } v j Internal Nodes from set {v 1,v 2,,v k-1 } v k v 1,v 2,,v k-1 v 1,v 2,,v k-1 Internal Nodes from set {v 1,v 2,,v k-1 } v i CASE-2 v j

50 Computing W k w ij [k] = w ij [k-1] V (w ik [k-1] Λ w kj [k-1] ) oper Procedure Warshall-transitive-closure(M R ) W=M R for k=1 to n begin for i=1 to n begin for j=1 to n w ij [k] = w ij [k-1] V (w ik [k-1] Λ w kj [k-1] ) end end end W is the answer M R + Complexity: (2n 2 )n=o(n 3 )

51 Equivalence Relation

52 Definition Three important characteristics of the notion equivalence : Every element is equivalent to itself (reflexivity) If a is equivalent to b, then b is equivalent to a (symmetry) If a is equivalent to b, and b is equivalent to c, then a is equivalent to c (transitivity) A binary relation R on a set A is an equivalence relation if R is reflexive, symmetric and transitive.

53 Modular equivalences: Congruence Modulo m R={(a,b) Reflexive as ara Symmetric: If arb=>m (a-b)=>(a-b)=km, where k is an integer Thus, (b-a)=-km=>m (b-a)=>bra Transitive: a b(mod m) arb=>(a-b)=k 1 m brc=>(b-c)=k 2 m So, (a-c)=(a-b)+(b-c)=(k 1 +k 2 )m=>m (a-c)=>arc

54 Equivalence Class Let R be an equivalence relation on a set A. The set of all the elements that are related to an element a of A is called the equivalence class of a. It is denoted by [a] R. When only one relation is under consideration, one can drop the subscript R. [a] R ={s (a,s)єr}. Any element in the class can be chosen as the representative element in the class.

55 Example arb iff a=b or a=-b R is an Equivalence relation (exercise) What is the equivalence class of an integer a? [a] R ={-a,a}

56 Example What are the equivalence classes of 0 and 1 for congruence modulo 4? [0]={,-8,-4,0,4,8, } [1]={,-7,-3,1,5,9, } The equivalence classes are called congruent classes modulo m.

57 Partitions Let R be an equivalence relation on a set A. These statements are equivalent if: 1. arb 2. [a]=[b] 3. [a] [b] Ø 1=>2=>3=>1

58 Theorem Let R be an equivalence relation on set A. 1. For, all a, bєa, either [a]=[b] or [a] [b]=ø 2. U xєa [x]=a Thus, the equivalence classes form a partition of A. By partition we mean a collection of disjoint nonempty subsets of A, that have A as their union.

59 Why both conditions 1 and 2 are required? In the class we had a discussion, saying that is 1 sufficient and does 2 always hold? Lets consider the following example: Define over the set A={y y ЄI + } R={(a,b) b=a 2 }. Thus (1,1),(2,4) are members of R. Consider the class: [x]={s (x,s)єr}

60 Pictographic Representation So, we see that we have classes which satisfy property 1 (here for distinct a and b, the intersection of [a] and [b] is always null) But the union of the partitions is not the set A. It s a subset of A For equivalence classes it is exactly A. Property 1 and 2 together define equivalence classes. [1]={1} [2]={4} [3]={9} [4]={16}

61 Quotient Set Let R be an equivalence relation on A. The quotient set, A/R, is the partition {[a] R aєa}. The quotient set is also called A modulo R or the partition of A induced by R. Equivalence classes of R form a partition of A. Conversely, given a partition {A i i ЄI} of A, there is an equivalence relation R that has the sets, A i as its equivalence classes. Equivalence relations induce partitions and partitions induce equivalence relations